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I was faced with the following question:

PART C – CHEMISTRY

Which one of the following properties is not shown by $\ce{NO}$?

  1. It combines with oxygen to form nitrogen dioxide

  2. Its bond order is $2.5$

  3. It is diamagnetic in the gaseous state

  4. It is a neutral oxide

I am confused about (3) & (4). I know $\ce{NO}$ reacts with water & oxygen to produce $\ce{HNO2}$, and then $\ce{NaOH}$ to produce $\ce{NaNO2, N2O, H2O}$. So, obviously it's acidic.

But there is one unpaired electron in the $\ce{N}$ atom. So, I see no reason for $\ce{NO}$ to be diamagnetic in gaseous state.

But, these questions do not have two correct answers. So, one of them has to be wrong. I am not sure which one.

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  1. $$\ce{2NO + O2 → 2NO2}$$ Nitric oxide reacts with oxygen to produce nitrogen dioxide. This reaction takes place quickly at room temperature in air. (chemiday)

  2. The bond order in $\ce{NO}$ is 2.5

  3. The oxidation state of $\ce N$ is +2, and this is a colourless, paramagnetic gas (ie. it is a radical, or contains an unpaired electron). (source)

  4. Neutral oxides are those oxides which show neither basic nor acidic properties when they react with water. Examples include carbon monoxide $(\ce{CO})$ and nitrous oxide $(\ce{N2O})$ which are only slightly soluble in water, and nitric oxide $\mathbf{(NO)}$ which is appreciably soluble in cold water. (source)

So, option 3 is the correct answer as it is an inaccurate statement.

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  • $\begingroup$ but NO reacts with NaOH. So, why isn't acidic? $\endgroup$ – Mockingbird May 8 '17 at 12:31
  • $\begingroup$ NO itself doesn't react with NaOH, so it is not acidic. It is only the downstream reaction products of NO with O2 that are acidic. $\endgroup$ – hBy2Py May 8 '17 at 13:15
  • $\begingroup$ Also, I take issue with that first citation. The oxidation of NO is second order in the NO concentration, so it is only a "quick" reaction when the concentration of NO is high. $\endgroup$ – hBy2Py May 8 '17 at 13:24
  • $\begingroup$ chemiday.com/en/reaction/3-1-0-8020 So is this wrong? $\endgroup$ – Mockingbird May 8 '17 at 13:43

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